67 lines
1.4 KiB
Markdown
67 lines
1.4 KiB
Markdown
# Na podstawie definicji obliczyć pochodną funkcji f w podanym punkcie $x_0$.
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## 1
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$f(x)=2x-x^{2},gdzie\ x_0=1$
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$f'(x)=2-2x$
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$f'(x_{0})=0$
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## 2
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$f(x)=x^{2}-7x,gdzie\ x_{0}=0$
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$f'(x)=2x-7$
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$f'(x_{0})=-7$
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## 3
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$f(x)=x^{3},x_{0}=1$
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$f'(x)=3x^2$
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$f'(x_{0})=3$
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## 4
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$f(x)=-2x^{3}+x, x_{0}=-1$
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$f'(x)=-6x^{2}+1$
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$f'(x_{0})=-5$
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## 5
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$f(x)=-\sqrt{x+2},\ x_{0}=2$
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$f(x)=-(x+2)^{\frac{1}{2}},\ x_{0}=2$
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$f'(x)=-\frac{1}{2\sqrt{x+2}}$
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$f'(x_{0})=-\frac{1}{4}$
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## 6
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$f(x)=\sqrt{1+2x},\ x_{0}=4$
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$f(x)=(1+2x)^{\frac{1}{2}}$
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$f'(x)=\cfrac{\frac{1}{2\sqrt{1+2x}}}{2}$
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$f'(x_{0})=\frac{1}{3}$
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## 7
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$f(x)=3+2\sqrt{4x-1},\ x_{0}=2\frac{1}{2}$
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$f'(x)=2\cdot \frac{1}{2\sqrt{4x-1}}\cdot 4$
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$f'(x_{0})=\frac{4}{3}$
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## 8
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$f(x)=\frac{1}{x^{2}},\ x_{0}=1$
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$f'(x)=-\frac{2}{x^3}$
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$f'(x_{0})=-2$
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## 9
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$f(x)=\frac{2}{x^{3}},\ x_{0}=-1$
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$f(x)=2(x^{-3})$
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$f'(x)=-\frac{6}{x^{4}}$
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$f'(x_{0})=-6$
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## 10
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$f(x)=\sqrt{x^{2}-1},\ x_{0}=-3$
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$f'(x)=\cfrac{1}{2\sqrt{x^{2}-1}} \cdot 2x$
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$f'(x_{0})=\frac{-6}{4\sqrt{2}}=\frac{-3}{2\sqrt{2}}$
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## 11
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$f(x)=\sqrt{2x^{2}+1},\ x_{0}=2$
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$f'(x)=\cfrac{1}{2\sqrt{2x^{2}+1}} \cdot 4x$
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$f'(x_{0})=\frac{4}{3}$
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## 12
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$f(x)=\sin{x},\ x_{0}=\frac{\pi}{2}$
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$f'(x)=\cos x$
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$f'(x_{0})=0$
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## 13
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$f(x)=\cos{x},\ x_{0}=\frac{\pi}{2}$
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$f'(x)=-\sin x$
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$f'(x_{0})=-1$
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## 14
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$f(x)=\sin{x^2},\ x_{0}=0$
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$f'(x)=\cos x^{2}\cdot 2x$
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$f'(x_{0})=0$
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## 15
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$f(x)=\sin\sqrt{x},\ x_{0}=0$
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$f'(x)=\cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}}$
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$f'(x_{0})=1\cdot\frac{1}{0}=brak\ rozwiązania$
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